Rotational Motion Question 129
Question: A thick walled hollow sphere has outer radius $R$ . It rolls down an inclined plane without slipping and its speed at the bottom is$v$ . If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom is $5v/4$ . What is the radius of gyration of the sphere?
Options:
A) $\frac{R}{\sqrt{2}}$
B) $\frac{R}{2}$
C) $\frac{3R}{4}$
D) $\frac{\sqrt{3}R}{4}$
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Case I: $mgh=\frac{1}{2}mv^{2}+\frac{1}{2}mk^{2}{{\omega }^{2}}$
$ Case II: $
$mgh=\frac{1}{2}m\times {{\left( \frac{5v}{4} \right)}^{2}}$
$\frac{1}{2}mk^{2}\frac{v^{2}}{R^{2}}+\frac{1}{2}mv^{2}=\frac{1}{2}m\times \frac{25v^{2}}{16}$
$\frac{k^{2}}{R^{2}}+1=\frac{25}{16}\Rightarrow k=\frac{3R}{4}$
